$$$\sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}$$$ 的积分

求$$$\int \sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}\, dx$$$。

使用公式 $$$\sin\left(\alpha \right)\sin\left(\beta \right)=\frac{1}{2} \cos\left(\alpha-\beta \right)-\frac{1}{2} \cos\left(\alpha+\beta \right)$$$，取 $$$\alpha=x$$$ 和 $$$\beta=2 x$$$，将 $$$\sin\left(x \right)\sin\left(2 x \right)$$$ 重写:

$${\color{red}{\int{\sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)} d x}}} = {\color{red}{\int{\left(\frac{\cos{\left(x \right)}}{2} - \frac{\cos{\left(3 x \right)}}{2}\right) \cos{\left(x \right)} d x}}}$$

展开该表达式:

$${\color{red}{\int{\left(\frac{\cos{\left(x \right)}}{2} - \frac{\cos{\left(3 x \right)}}{2}\right) \cos{\left(x \right)} d x}}} = {\color{red}{\int{\left(\frac{\cos^{2}{\left(x \right)}}{2} - \frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{2}\right)d x}}}$$

对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(x \right)} = \cos^{2}{\left(x \right)} - \cos{\left(x \right)} \cos{\left(3 x \right)}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$${\color{red}{\int{\left(\frac{\cos^{2}{\left(x \right)}}{2} - \frac{\cos{\left(x \right)} \cos{\left(3 x \right)}}{2}\right)d x}}} = {\color{red}{\left(\frac{\int{\left(\cos^{2}{\left(x \right)} - \cos{\left(x \right)} \cos{\left(3 x \right)}\right)d x}}{2}\right)}}$$

逐项积分：

$$\frac{{\color{red}{\int{\left(\cos^{2}{\left(x \right)} - \cos{\left(x \right)} \cos{\left(3 x \right)}\right)d x}}}}{2} = \frac{{\color{red}{\left(- \int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x} + \int{\cos^{2}{\left(x \right)} d x}\right)}}}{2}$$

应用降幂公式 $$$\cos^{2}{\left(\alpha \right)} = \frac{\cos{\left(2 \alpha \right)}}{2} + \frac{1}{2}$$$，并令 $$$\alpha=x$$$:

$$- \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{{\color{red}{\int{\cos^{2}{\left(x \right)} d x}}}}{2} = - \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{{\color{red}{\int{\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)d x}}}}{2}$$

对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(x \right)} = \cos{\left(2 x \right)} + 1$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$$- \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{{\color{red}{\int{\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{1}{2}\right)d x}}}}{2} = - \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{{\color{red}{\left(\frac{\int{\left(\cos{\left(2 x \right)} + 1\right)d x}}{2}\right)}}}{2}$$

逐项积分：

$$- \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{{\color{red}{\int{\left(\cos{\left(2 x \right)} + 1\right)d x}}}}{4} = - \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{{\color{red}{\left(\int{1 d x} + \int{\cos{\left(2 x \right)} d x}\right)}}}{4}$$

应用常数法则 $$$\int c\, dx = c x$$$，使用 $$$c=1$$$：

$$- \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{\int{\cos{\left(2 x \right)} d x}}{4} + \frac{{\color{red}{\int{1 d x}}}}{4} = - \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{\int{\cos{\left(2 x \right)} d x}}{4} + \frac{{\color{red}{x}}}{4}$$

设$$$u=2 x$$$。

则$$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (步骤见»)，并有$$$dx = \frac{du}{2}$$$。

所以，

$$\frac{x}{4} - \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{{\color{red}{\int{\cos{\left(2 x \right)} d x}}}}{4} = \frac{x}{4} - \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{4}$$

对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(u \right)} = \cos{\left(u \right)}$$$ 应用常数倍法则 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$：

$$\frac{x}{4} - \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{{\color{red}{\int{\frac{\cos{\left(u \right)}}{2} d u}}}}{4} = \frac{x}{4} - \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{{\color{red}{\left(\frac{\int{\cos{\left(u \right)} d u}}{2}\right)}}}{4}$$

余弦函数的积分为 $$$\int{\cos{\left(u \right)} d u} = \sin{\left(u \right)}$$$：

$$\frac{x}{4} - \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{{\color{red}{\int{\cos{\left(u \right)} d u}}}}{8} = \frac{x}{4} - \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{{\color{red}{\sin{\left(u \right)}}}}{8}$$

回忆一下 $$$u=2 x$$$:

$$\frac{x}{4} - \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{\sin{\left({\color{red}{u}} \right)}}{8} = \frac{x}{4} - \frac{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}{2} + \frac{\sin{\left({\color{red}{\left(2 x\right)}} \right)}}{8}$$

使用公式 $$$\cos\left(\alpha \right)\cos\left(\beta \right)=\frac{1}{2} \cos\left(\alpha-\beta \right)+\frac{1}{2} \cos\left(\alpha+\beta \right)$$$ 并结合 $$$\alpha=x$$$ 和 $$$\beta=3 x$$$ 重写被积函数:

$$\frac{x}{4} + \frac{\sin{\left(2 x \right)}}{8} - \frac{{\color{red}{\int{\cos{\left(x \right)} \cos{\left(3 x \right)} d x}}}}{2} = \frac{x}{4} + \frac{\sin{\left(2 x \right)}}{8} - \frac{{\color{red}{\int{\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{\cos{\left(4 x \right)}}{2}\right)d x}}}}{2}$$

对 $$$c=\frac{1}{2}$$$ 和 $$$f{\left(x \right)} = \cos{\left(2 x \right)} + \cos{\left(4 x \right)}$$$ 应用常数倍法则 $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$：

$$\frac{x}{4} + \frac{\sin{\left(2 x \right)}}{8} - \frac{{\color{red}{\int{\left(\frac{\cos{\left(2 x \right)}}{2} + \frac{\cos{\left(4 x \right)}}{2}\right)d x}}}}{2} = \frac{x}{4} + \frac{\sin{\left(2 x \right)}}{8} - \frac{{\color{red}{\left(\frac{\int{\left(\cos{\left(2 x \right)} + \cos{\left(4 x \right)}\right)d x}}{2}\right)}}}{2}$$

逐项积分：

$$\frac{x}{4} + \frac{\sin{\left(2 x \right)}}{8} - \frac{{\color{red}{\int{\left(\cos{\left(2 x \right)} + \cos{\left(4 x \right)}\right)d x}}}}{4} = \frac{x}{4} + \frac{\sin{\left(2 x \right)}}{8} - \frac{{\color{red}{\left(\int{\cos{\left(2 x \right)} d x} + \int{\cos{\left(4 x \right)} d x}\right)}}}{4}$$

积分 $$$\int{\cos{\left(2 x \right)} d x}$$$ 已经计算过：

$$\int{\cos{\left(2 x \right)} d x} = \frac{\sin{\left(2 x \right)}}{2}$$

因此，

$$\frac{x}{4} + \frac{\sin{\left(2 x \right)}}{8} - \frac{\int{\cos{\left(4 x \right)} d x}}{4} - \frac{{\color{red}{\int{\cos{\left(2 x \right)} d x}}}}{4} = \frac{x}{4} + \frac{\sin{\left(2 x \right)}}{8} - \frac{\int{\cos{\left(4 x \right)} d x}}{4} - \frac{{\color{red}{\left(\frac{\sin{\left(2 x \right)}}{2}\right)}}}{4}$$

设$$$v=4 x$$$。

则$$$dv=\left(4 x\right)^{\prime }dx = 4 dx$$$ (步骤见»)，并有$$$dx = \frac{dv}{4}$$$。

该积分可以改写为

$$\frac{x}{4} - \frac{{\color{red}{\int{\cos{\left(4 x \right)} d x}}}}{4} = \frac{x}{4} - \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{4} d v}}}}{4}$$

对 $$$c=\frac{1}{4}$$$ 和 $$$f{\left(v \right)} = \cos{\left(v \right)}$$$ 应用常数倍法则 $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$：

$$\frac{x}{4} - \frac{{\color{red}{\int{\frac{\cos{\left(v \right)}}{4} d v}}}}{4} = \frac{x}{4} - \frac{{\color{red}{\left(\frac{\int{\cos{\left(v \right)} d v}}{4}\right)}}}{4}$$

余弦函数的积分为 $$$\int{\cos{\left(v \right)} d v} = \sin{\left(v \right)}$$$：

$$\frac{x}{4} - \frac{{\color{red}{\int{\cos{\left(v \right)} d v}}}}{16} = \frac{x}{4} - \frac{{\color{red}{\sin{\left(v \right)}}}}{16}$$

回忆一下 $$$v=4 x$$$:

$$\frac{x}{4} - \frac{\sin{\left({\color{red}{v}} \right)}}{16} = \frac{x}{4} - \frac{\sin{\left({\color{red}{\left(4 x\right)}} \right)}}{16}$$

因此，

$$\int{\sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)} d x} = \frac{x}{4} - \frac{\sin{\left(4 x \right)}}{16}$$

加上积分常数：

$$\int{\sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)} d x} = \frac{x}{4} - \frac{\sin{\left(4 x \right)}}{16}+C$$

$$$\int \sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)}\, dx = \left(\frac{x}{4} - \frac{\sin{\left(4 x \right)}}{16}\right) + C$$$A